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Carbonyl Compound Lab

Reactions of Carbonyl Compounds

Organic Chemistry II Lab

April 6, 2000

Physical Constants:

 

SUBSTANCE FORMULA MOL. MASS (g) MELTING POINT (oC) BOILING POINT (oC) DENSITY

(g/ml)

Water H2O 18 0.0 100 1.0
Sulfuric Acid H2SO4 98.08 10.31 337 1.8
Methanol CH3OH 32.04 -97.6 64.6 0.79
Sodium Acetate NaC2H3O2 82.04 324 1.53

Results:

 

UNKNOWN DNP TEST Semicarbazone

 

Possible identities

(based on MP)

COLOR MELTING  POINT (oC) Color Melting Point (oC)
A Orange-Yellow 228 Snowy White 216 p-Tolualdehyde

Benzaldehyde

B  

Orange-Red

 

154 Snowy White 164 Acetaldehyde

Cyclohexanone

C No percipitate NA NA NA Not an aldehyde or ketone

 

 

Discussion:

UNKNOWN A: Figure 1 represents compound C.  Based on the melting points of the DNP precipitate and the semicarbazone derivative, the possibilities were brought down to salicylaldehyde, benzaldehyde, p-tolualdehyde.  As mentioned beforehand, there is a strong peak at 1600 in the IR that represents benzene rings and the two peaks at 690 and 750 are characteristic of monosubstituted benzene rings.  Of the three possibilities, the only one that is a monosubstituted benzene ring is benzaldehyde.  Also analysis of the NMR showed that a peak at 6.6 ppm with a 10.0 shift a carbonyl group which is present in aldehydes.  Other peaks between 7.3-7.8 ppm indicated the presence of aromatic hydrogens.  This all indicates that p-tolualdehyde can’t be unknown A since p-toualdehyde has a low chemical shift found between 1-2 ppm.  Therefore, compound A can easily be determined as BENZALDEHYDE.

 

UNKNOWN B: Based on the experimental melting points obtained of the DNP (154oC) precipitate, which was positive, and the semicarbazone derivative (164oC), the possibilities were narrowed down to acetaldehyde and cyclohexanone (by comparison with table of known melting points).  All other compounds differed greatly in their m.p. values.  They were only close with one of the two melting points or were lower than the experimental values.  Usually, the experimental values of melting points are lower than the values of the pure substances due to wetness and both possibility chosen have slightly higher melting points.  To make the final decision, the IR and NMR of the two were compared to one of the three figures.  Figure three was easily eliminated because it shows an IR for an alcohol. It could be said that it represented acetylaldehyde on the basis of the medium peak (IR) at around 2800 cm-1, indicating the presence of C-H bonds.  However, at 1600 cm-1, there is a band that represents benzene rings. It was seen that a strong band in the IR at around 2900 exists, which also represents C-H bonds.  This could be a possibility for cyclohexanone since it has 10 C-H bonds.  The NMR was affirm this.  A ketone has 2 close peaks on the NMR.  One peak was present at 2 ppm implying ketone.  Also the multiplet of the peak may mean a ring.  Therefore, it could be concluded with relatively high likeliness that compound B was CYCLOHEXANONE.

 

UNKNOWN C: Based on the experimental results (negative DNP test), it could be concluded that this compound was neither an aldehyde nor a ketone.  The analysis had to be solely relied upon the IR spectra and the NMR spectra.  It had already been deduced that the corresponding data for compound C was figure 3.  From preliminary observations, it was concluded that the substance was an alcohol (strong and broad IR peak at around wavelength 3 microns) and did not contain a benzene ring (based on molecular formula – C4H10O).  It is known that a double bond does not exist in the compound because there is no IR band at 1620-1680 cm-1.  The first peak observed on the NMR graph is at about 4 ppm and it is a singlet.  It is also the furthest shift to the left, which usually signifies proton binding to or near an electronegative element, O.  The second peak, observed at around 3.5 ppm, is a doublet and shows that there is a single proton next to it.  The third peak, a multiplet, is splitted many times due to its many different neighbors.  The final signal is split into a doublet, but has higher intensity, indicating that there is more than one of these kinds of protons.  Through these stated observations we concluded that unknown A was 2 methyl-propanol.

 

 

Questions:

 

1.         If benzyl alcohol is oxidized by O2 it would form benzaldehyde and yield a precipitate in the DNP test.  In contrast to this, benzyl alcohol does not yield a precipitate.

 

2. Aldehydes and ketones give positive results for DNP tests. This is because of the carbonyl nature of these compounds.  There would have to be a band near 1720-1740, 1705-1725, or 1680-1700 in their IR spectra.  One of these bands would indicate a positive DNP test.  A will give a precipitate during the DNP test.

 

3 In an IR spectrum, peaks that occur in the range between 730-770 and 690 to 710 represent aromatic rings.  Thus, we can conclude that compounds A, B, C and D are all aromatic compounds.

4. A structure for compound C based on the IR/NMR spectra would be:

 

 

 

 

5.      Compound D is 1-Phenylethanol.  We can tell this because in the IR spectrum, there is a peak in the range between 3200-3600, which represents an alcohol.  The peak at 7 ppm represents a ring structure.  The peak at 5.1ppm is split into a quartet.  The peak at 4ppm is a singlet.  And the peak at 1.4 ppm is a doublet.  This splitting is characteristic of 1-Phenylethanol.

 

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